Question

find the greates number which divides 6168,2447 and 3118 leaving the same reminder in each case?

Answer 1

Since the remainders are same in each division, the difference of numbers will be exactly divisible by the 'greatest number'

Answer 2

so, the required 'greatest divisor' wud be : \(gcf(6168-2447, 6118-3118, 3118-2447)\)

Answer 3

dear sometime we only leave reminder and take GCF....wy in this case we a r taking diff

Answer 4

good question :)
in those cases where u directly take gcf, you're trying to find gcf of given numbers.

but in the present problem, you're trying to find 'gcf of difference of numbers'

Answer 5

below two phrases are same :-
1) gcf of difference of given set of numbers
2) greatest number that leaves the same remainder when it divides the given of set numbers

Answer 6

why they both are same ?

Answer 7

lets dive in further

Answer 8

like in this case we only take GCF after leaving reminder
find the greatest number which divides 742 and 1162 leaving 7 as reminder in each case?

Answer 9

taking gcf gives u the 'greatest number that leaves 0 as remainder'

Answer 10

why we take diff in 3 numbers?

Answer 11

before we address that, lets finish ur leaving 7 as remainder problem :)

Answer 12

find the greatest number which divides 742 and 1162 leaving 7 as reminder in each case?

you seem to be knowing how to work this problem... may i knw how u solve this ?

Answer 13

occording question i leave 7 in each number and than GCF

Answer 14

you mean, u take \(gcf(742-7, 1162-7)\) ?

Answer 15

yes

Answer 16

gcf(735, 1155)

105

Answer 17

and u claim that, 742/105 and 1162/105 leave the remainder 7

Answer 18

which is correct !

Answer 19

may i knw why this works ?

Answer 20

yes correct

Answer 21

dnt know

Answer 22

okay, there is a standard way for working these problems

Answer 23

if you're not formula oriented, you will like the standard way

Answer 24

ok

Answer 25

find the greatest number which divides 742 and 1162 leaving 7 as reminder in each case?

say, the required 'greatest number is \(n\)'

Answer 26

since when \(n\) divides 742, it leaves a remainder 7 :-
742 = \(n\)k + 7

742 - 7 = \(n\)k

Answer 27

ok

Answer 28

similarly,
since when \(n\) divides 1162, it leaves a remaidner 7 :-
1162 = \(nq + 7\)

1162-7 = \(nq\)

Answer 29

ok

Answer 30

from both above, 742-7 and 1162-7 are divisible by \(n\) (why ?)

Answer 31

yes these are divisible

Answer 32

so, taking the \(gcf(742-7, 1162-7)\) gives u the 'largest number' that leaves remainder 7

Answer 33

see if that seems convincing :)

Answer 34

now i understand

Answer 35

good, see if u can explain why we're taking the differences now (our original problem)

Answer 36

u may use similar reasoning to make sense of it

Answer 37

find the greates number which divides 6168,2447 and 3118 leaving the same reminder in each case?

Answer 38

right

Answer 39

when our 'greatest number', say \(n\) divides the given numbers, say, it leaves a remainder \(R\) :-

6168 = \(n\)p + \(R\)
2447= \(n\)q + \(R\)
3118= \(n\)r + \(R\)

Answer 40

clearly, when u subtract any two equations above, \(R\) cancels out, and the difference is divisible by \(n\)

Answer 41

So, the required 'greatest number' is simple the gcf of difference of given numbers

Answer 42

thanks Dear

Answer 43

np :)